= Lazard isomorphism
{c}
{title2=$H_c^*(G,M)\cong H^*(\mathfrak g,M)$}
The <Lazard isomorphism> compares <continuous group cohomology> of suitable $p$-adic analytic <groups> with <Lie algebra cohomology> of their associated <Lie algebras>. It converts a cohomological question about a nonlinear multiplication law into a calculation with alternating multilinear maps. The hypotheses matter: one cannot assert this comparison for every <topological group>.
One precise integral version takes a <uniform pro-p group> $G$ and a finite free $\mathbb Z_p$-<module> $M$. For odd $p$, require the continuous action to have image in $1+p\operatorname{End}(M)$; for $p=2$, use $1+4\operatorname{End}(M)$. With $\mathfrak g$ the associated integral <Lazard Lie algebra>,
$$
H_c^*(G,M)\cong H^*(\mathfrak g,M).
$$
This is an integral refinement of the rational comparison.<Some complements to the Lazard isomorphism by Annette Huber and others>[Theorem 3.3.3, printed manuscript page 16]{ref}
For the additive <group> $G=\mathbb Z_p$ with trivial coefficients $M=\mathbb Z_p$, continuous one-cocycles are $x\mapsto ax$. The associated rank-one abelian <Lie algebra> has $H^1\cong\operatorname{Hom}_{\mathbb Z_p}(\mathbb Z_p,\mathbb Z_p)\cong\mathbb Z_p$, giving the same answer. Rationalizing coefficients alone would not establish an integral comparison in general.
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